A Triangle and Two Regular Convex Polygons

Consider the regular triangle, $n_1$ -gon, and $n_2$ -gon. The sum of their internal angles is $\left(180^\circ - \frac{360^\circ}{3}\right) + \left(180^\circ - \frac{360^\circ}{n_1}\right) + \left(180^\circ - \frac{360^\circ}{n_2}\right) = 420^\circ - \frac{360^\circ}{n_1} - \frac{360^\circ}{n_2}.$ This must be equal to $360^\circ$ ; thus we have $\frac{360^\circ}{n_1} + \frac{360^\circ}{n_2} = 60^\circ.$ Divide by $60^\circ$ and multiply by $n_1n_2$ to find $6(n_1 + n_2) = n_1n_2\ \ \ \ \ \therefore\ \ \ \ \ \ n_1n_2 - 6(n_1 + n_2) = 0.$ Complete the product: $(n_1 - 6)(n_2 - 6) = 36.$ The factoring of 36 with the highest factor is $1\cdot 36 = 36$ , so that $n_1 - 6 = 1$ and $n_2 - 6 = 36$ . This shows that $n_1 = 7$ and $n_2 = \boxed{42}$ .

The polygon with fewer sides must have an exterior angle less than 60 degrees (the angle of the equilateral triangle), and the closer it is to 60 the smaller the contribution from the second polygon, and so the more sides it will have.

A hexagon has an external angle of exactly 60 degrees, so the first polygon must have seven sides with external angle $360/7$ (about 51.4...)

The second polygon has an external angle of $60 - 360/7$ (8.57...) so the number of sides is $360 / (60 - 360/7)$ = 42

Let $m$ be the number of sides of one of the regular convex polygons, and $n$ be the number of sides of the other regular convex polygon. Then each of their interior angles are $\frac{\pi(m - 2)}{m}$ and $\frac{\pi(n - 2)}{n}$ , respectively, and the interior angle of the equilateral triangle is $\frac{\pi}{3}$ . In the given problem, all three of these interior angles must add exactly to the space surrounding a point, which is $2\pi$ , and so $\frac{\pi(m - 2)}{m} + \frac{\pi(n - 2)}{n} + \frac{\pi}{3} = 2\pi$ , and this simplifies to $m = \frac{6n}{n - 6}$ .

Since $m = \frac{6n}{n - 6}$ , $m > 0$ only when $n > 6$ . Therefore, since $m > 3$ (since $m$ represents a side of a regular convex polygon), $n > 6$ . Also, since $m’ = \frac{-36}{(n - 6)^2} < 0$ , it is always decreasing, which means that the maximum possible $m$ value occurs at the first possible $n$ value after $n > 6$ , and since $n$ is an integer (since $n$ represents a side of a regular convex polygon), this is when $n = 7$ . Therefore, the largest number of sides is $m = \frac{6 \cdot 7}{7 - 6} = \boxed{42}$ .

The 3 internal angles sum to 360deg. The two unknown polygons are an m-gon and an n-gon.

(180m-360)/m + (180n-360)/n + 60 = 360 in degrees.

Boils down to mn - 6m = 6n m = 6 + 36/(n-6) The highest n such that m is an integer is n=42. m=7.

Answer is 42 sides.

Consider the following implications :-

1. The sum of interior angles of the two polygons is 360 -60 = 300 degrees.

2. No interior angle can be greater than or even equal to 180, as minimum possible exterior angle for a polygon (with infinite sides) is 0 degrees, as $\frac{360}{infinite no. of sides}$ = 0

Therefore, The two conditions for two respective interior angles A and B are
A + B = 300 degrees A , B < 180 degrees Let us suppose one of them is 180 degrees, ie A = 180 degrees, then B = 300 - 180 = 120 degrees.

Interior angle for a polygon with n no. of sides = $\frac{180 * (n - 2)}{ n }$ = 120 , which yields n = 6 Therefore, when no of sides of one polygon is maximum, the minimum no. of sides for another polygon can only be 6. But we cannot have infinite sides on any polygon, so

Let us take the next minimum number of sides on the other polygon , 7; Interior angle of a polygon with 7 sides = $\frac{180 * 5}{7}$ , hence the interior angle for the second maximum no of sides of a polygon would be = 300 - 180 $\frac{5}{7}$

= $\frac{2100 - 900}{7}$

= $\frac{1200}{7}$

Now, $\frac{180 * (n - 2)}{ n }$ = $\frac{1200}{7}$

Therefore, $\frac{n - 2}{n}$ = $\frac{20}{21}$ , which yields n = 42

As the interior angles of a regular polygon (the interior angle represented by a) must fit: a < 180 Then we must find the largest interior angle shape. We look through what it can be, and the first polygon that fits is a heptagon, with its interior angles being $\frac{(7-2)*180}{7}$

From here, we know that the current angle total we have is about 188.57142 degrees.

We subtract this from 360, giving 171.42857 degrees per each interior angle of this shape.

To find the number of sides this shape has, we first figure out the exterior angles, by doing 180-171.42857=8.57143

Since exterior angles add up to 360, we can figure out how many exterior angles there are, by doing $\frac{360}{8.57143}$

This gives us 42 exterior angles, which is by extension 42 sides.

The external angles (i.e. 180 - internal angle) of the polygons have to sum to 60 degrees i.e. internal angle of the triangle. The external angle of an n sided polygon is 360/n degrees. Hence $\frac{360}{n}$ + $\frac{360}{m}$ =60 where n and m are integers representing the number of sides of the two polygons. Thus $\frac{1}{n}$ + $\frac{1}{m}$ = $\frac{1}{6}$ . Clearly n and m MUST both be larger than 6 for this equation to hold. So if n=7 then m=42. If n=8 then m=24 all the way to n=42 and m=7. So as one number increases, the other decreases. Hence the largest polygon will have n or m =42 sides.

N1 = 6 and N2 = infinity works too.

We have two regular polygones and a triangle, where the contribution of their angles are

$60^\circ + (180^\circ - 360^\circ /n_1) + (180^\circ - 360^\circ / n_2) = 360^\circ .$

It is clear that, when $n_1$ increases, then $n_2$ decreases. Moreover, they follow

$n_1 = \dfrac{n_2}{\tfrac{1}{6}n_2 -1}.$

The limiting case is when $n_1 \rightarrow \infty$ , i.e. a circle, in that case, the sum of the angles would be $60^\circ + 180^\circ + 240^\circ = 360^\circ$ , where $240^\circ$ corresponds to $n_2 = 6$ . Therefore, the solution is the pair corresponding to the next integer, i.e. $n_2 = 7$ and $n_1 = 42$ ; the sum of the angles would be $60^\circ + 171.43^\circ + 128.57^\circ = 360^\circ$ .

The total angle of the two remaining polygons must be 300. No shape can have an angle greater than or equal to 180 (without having an infinite number sides which is impossible). Therefore we can deduce that the shape with the fewest number of sides must have more than 6 sides (as a shape with 6 sides has angles of 120 degrees which would leave 180 degrees remaining)

We have already been shown that an 8 side shape will work so we only have to test with a 7 sided shape.

The angle in a heptagon is roughly 128.5. Therefore the other shape must have angles of 300-128.5= 171.5 degrees The total angle of this shape is therefore the number of sides (n) multiplied by 171.5 However we also know that the total interior angle of a polygon is 180(n-2)

Set these equations as equal to each other and solve 171.5n = 180(n-2)

171.5n = 180n - 360

360 = 8.5n

42.3...= n

Therefore the larger number of sides is 42